Question: Let $h(x)=\begin{cases} e^{2x}&\text{for }x<0 \\\\ e^{5x}&\text{for }x\geq 0 \end{cases}$ Is $h$ continuous at $x=0$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Solution: For $h$ to be continuous at $x=0$, we need $\lim_{x\to 0}h(x)$ and $h(0)$ to exist and be equal. Since $0\geq 0$, the rule that applies to $x=0$ is $e^{5x}$. So $h(0)=e^{5(0)}=1$. Now let's analyze $\lim_{x\to 0}h(x)$. Finding $\lim_{x\to 0^{ +}}h(x)$ For $x$ -values larger than $0$, the appropriate rule for $h(x)$ is $e^{5x}$. Since $e^{5x}$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 0^{ +}}h(x) \\\\ &=\lim_{x\to 0^{ +}}[e^{5x}] \gray{e^{5x}\text{ is the rule for }x>0} \\\\ &=e^{5(0)} \gray{e^{5x}\text{ is continuous at }x=0} \\\\ &=1 \end{aligned}$ Finding $\lim_{x\to 0^{ -}}h(x)$ For $x$ -values smaller than $0$, the appropriate rule for $h(x)$ is $e^{2x}$. Since $e^{2x}$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 0^{ -}}h(x) \\\\ &=\lim_{x\to 0^{ -}}[e^{2x}] \gray{e^{2x}\text{ is the rule for }x<0} \\\\ &=e^{2(0)} \gray{e^{2x}\text{ is continuous at }x=0} \\\\ &=1 \end{aligned}$ Conclusion We found that: $\lim_{x\to 0^{ +}}h(x)=\lim_{x\to 0^{ -}}h(x)=h(0)=1$ Since the one-sided limits are both equal to $h(0)$, we can determine that the two-sided limit $\lim_{x\to 0}h(x)$ is also equal to $h(0)$, and $h$ is continuous at $x=0$.